Question

Write the chemical equation representing the reaction between Mg metal and sulfuric acid. Calculate the mass of hydrogen gas at zero degrees Celsius and 14.00 psi when 15 g of Magnesium reacts with 25 g Sulfuric Acid?

Answer 1

Write the balanced chemical equation.

Mg (s) + H₂SO₄ (aq) --------> MgSO₄ (aq) + H₂ (g)

As per the stoichiometric equation,

1 mole Mg = 1 mole H₂SO₄ = 1 mole H₂.

Atomic mass of Mg = 24.305 g/mol.

Molar mass of H₂SO₄ = (2*1.008 + 1*32.065 + 4*15.9994) g/mol = 98.0786 g/mol.

Molar mass of H₂ = (2*1.008) g/mol = 2.016 g/mol.

Find out the mole(s) of Mg and H₂SO₄ corresponding to 15 g Mg and 25 g H₂SO₄.

Mole(s) Mg = (15 g)/(24.305 g/mol) = 0.6171 mole.

Mole(s) H₂SO₄ = (25 g)/(98.0786 g/mol) = 0.2549 mole.

Offcourse, we see that H₂SO₄ is the limiting reactant and the yield of H₂ will be decided by the mole(s) of H₂SO₄.

Mole(s) H₂ produced = mole(s) H₂SO₄ used = 0.2549 mole.

Mass of H₂ produced = (0.2549 mole)*(2.016 g/mol) = 0.5139 g (ans).